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BM267 1
BM267 - Introduction to Data
Structures
Ankara University
Computer Engineering DepartmentBulent Tugrul
5. Trees
BM267 2
Learn about the definitions, characteristics and
implementation details for:
• General trees
• Rooted trees
• Binary and N-ary trees
• Tree operations, tree traversal algorithms
Objectives
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Tree Structures
One of the most frequently used ordering methods of data.
Many logical organizations of everyday data exhibit tree
structures
Promotional tournaments
Organizational charts
Hierarchical organization of entities
Parsing natural and computer languages
Game trees
Decision trees
...
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Trees
A real tree
Computer Scientist's tree
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Trees
A general tree is a nonempty collection of vertices
(nodes) and connections between nodes (edges) that
satisfy certain rules. These rules impose a hierarchical
structure on the nodes with a parent-child relation.
There is only one connecting path between any two
nodes.
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14
17 11
9
13
53 4
19 7
Trees
Root
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-
+ *
9 53 4 7
Trees
= ( 9 + 53 ) – ( 4 * 7 )
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Rooted Trees
• There is a unique node called root node. Node “14” is the root
of the tree.
• The parent of a node is the node linked above it. Every non-
root node has a unique parent. Node 17 is the parent of 9 and
53.
• The nodes whose parent is node n are n’s children. The
children of Node 17 are 9 and 53.
• Nodes without children are leaves. Nodes 13, 53, 19, and 7 are
leaves.
• Two nodes are siblings if the have the same parent. 9 and 53 are
siblings of each other.
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Root
Root’s
children
Leaves
Rooted Trees
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• An empty tree has no nodes
• The descendants of a node are its children and the descendents of its children
• The ancestors of a node are its parent (if any) and the ancestors of its parent
• An ordered tree is one in which the order of the children is important; an unordered tree is one in which the ordering of the children is not important.
• The branching factor of a node is the number of children it has.
Rooted Trees
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0
1
3
2
1 1 1
3
2 2 2 2 2 2
The depth or level of a node n is the number
of edges on a path from the root to n.
The depth of the root is 0.
Root is at level 0.
Rooted Trees
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Rooted Trees
0 0
0 0 0000 1
11 12
3
The height of a node n is the number of
edges on the longest path from n to a
descendent leaf.
The height of each leaf is 0.
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Binary Trees
A binary tree is a special rooted tree in which
every node has at most 2 children.
Children are ordered: every child is explicitly
designated as left or right child.
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Binary Trees
• The i-th level of a binary tree contains all nodes at
depth i.
• The height of a binary tree is the height of its root.
• The i-th level of a binary tree contains at most 2i
nodes.
• A binary tree of height h contains at most 2h+1–1
nodes.
• A binary tree of height h has at most 2h leaves.
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Binary Trees
Level 3
Level 1
Level 2
Level 0
23
21
22
20
Total nodes = 2h + 2h-1 +…+ 22 +21 +20
2h+1 - 1
= ------------
2 -1
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A binary tree is complete( perfect ) if:
• Every node has either zero or two children. (Every
internal node has two children.)
• Every leaf is at the same level.
Binary Trees
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Binary Trees
A binary tree is almost complete (perfect) if
• All levels of the tree are complete, except possibly
the last one.
• The nodes on the last level are as far left as possible.
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Binary Trees
• A almost complete binary tree of height h
contains between 2h and 2h + 1 – 1 nodes.
• A almost complete binary tree of size n has
height h = floor( log n ).
2h <= n <= 2h + 1 – 1
h <= log n < h+1
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Binary Trees
l g o r i t h m sa
i
m
o t
l
h
a
s
g
r
0
1 2
3 4 5 6
987
parent(i)= (i –1) / 2
left(i) = 2i + 1
right(i) = 2i + 2
(integer
division )
0 1 2 3 4 5 6 7 8 9
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Binary Trees
We can also represent incomplete binary trees in an array
A
B
C
0
21
6543
A B C
0 1 2 3 4 5 6
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Binary Trees
Linked representations of binary trees.
A
root
B G
E K M
struct TreeNode{
char data;
struct TreeNode *left;
struct TreeNode *right;
}
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Binary Trees
Common Binary Tree Operations
– Determine its height
– Determine the number of elements in it
– Display the binary tree on the screen.
Returns the height of the tree.
int height(link h)
{ int u, v;
if (h == NULL)
return -1;
u = height(h->l);
v = height(h->r);
if (u > v) return u+1;
else return v+1; }
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Returns the number of elements in the tree.
int count(link h){
if (h == NULL)
return 0;
return count(h->l) + count(h->r) + 1;
}
Binary Trees
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• To traverse (or walk) the binary tree is to visit each
node in the binary tree exactly once
• Tree traversals are naturally recursive.
• Since a binary tree has two dimensions, there are
two possible ways to traverse the binary tree
• Depth-first - visit nodes on the same path first
(start from top, go as far down as possible)
• Breadth-first - visit nodes at the same level first
(start from left, go as far right as possible)
Tree Traversals
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• Since a binary tree has three “parts,” there are three
possible ways to traverse the binary tree (from left to
right) :
• Pre-order: the node is visited first, then the
children (left to right)
• In-order: the left child is visited, then the
node, then the right child
• Post-order: the node is visited after the
children
Depth-first Traversals (binary trees)
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a
c
d e
j
i
h
g
f
b
Pre-order Traversal
h j d i c b g f a e
: Node is visited here
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a
c
d e
j
i
h
g
f
b
In-order Traversal
i d c j g b h a f e
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a
c
d e
j
i
h
g
f
b
Post-order Traversal
i c d g b j a e f h
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a
c
d e
j
i
h
g
f
b
Breadth-first Traversal
h j f d b a e i c g
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Prints the nodes’ data in Preorder
void traverse( LINK h )
{
if (h)
{
printf(“%d”, h->data); //(prints the node)
traverse(h->left);
traverse(h->right);
}
}
Tree Traversal - Preorder
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Tree Traversal - InorderPrints the nodes’ data in Inorder
void traverse( LINK h )
{
if (h)
{
traverse(h->left);
printf(“%d”, h->data); //(prints the node)
traverse(h->right);
}
}
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Tree Traversal - PostorderPrints the nodes’ data in Postorder
void traverse( LINK h )
{
if (h)
{
traverse(h->left);
traverse(h->right);
printf(“%d”, h->data); //(prints the node)
}
}
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Implementing (general) rooted trees
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Example: Variable-length codes
ASCII uses 8-bits for coding letters (fixed-length code).
To minimize the space requirements, we can use an
alternate coding scheme (variable-length code):
• Let the most frequently used letters be represented
with shorter bit sequences (depends on the
language being coded).
• Let the least frequently used letters be represented
with longer bit sequences.
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Requirements: For each possible coded sequence, the
sequence must be
• uniquely decodeable.
• instantaneously decodeable (without the need for
further computations or table look-ups).
This philosophy had been employed in Morse code.
Also known as Huffman coding.
Example: Variable-length codes
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Example: Variable-length codes
Let our alphabet consist of 5 symbols, A, B, C, D, E.
Consider the code for
ABCDE.
Symbol Freq.(%)
A 40
B 25
C 15
D 15
E 5
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Example: Variable-length codes
Assume the following
codes were chosen:
Symbol Code
A 1
B 00
C 01
D 11
E 011
Consider the coding for ABCDE.
The code will be: 1000111011
Can you decode it?
This code is not
uniquely decodeable.
1 . 00 . 01 . 11 ? 011
Is 011 = 011 (E) or 01.1 (CA) ?
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Example: Variable-length codes
Assume the following
codes were chosen:
Symbol Code
A 0
B 01
C 011
D 0111
E 111
Consider the coding for ABCDE.
The code will be: 0010110111111
Can you decode it?
This code is not instantaneously
decodeable. You have check the
next digit.
0.01 ? 0110111111
Is 011 = 01 (B) . 1 or 011 (C) ?
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Example: Variable-length codes
1
1
1
0
1
1
EC
B
A
1D
This code is not instantaneously
decodeable. You have check the
next digit (compare the next digit
with the next edge on the tree).
However, it is uniquely
decodeable.
Draw the code tree. Start from the root and follow
the edges until a code word is found.
Repeat until decoding is completed.
BM267 40
5
E
25
B
15
D
40
A
15
C
Huffman Coding
20
5
E
15
D
15
C
25
B
40
A
Initial State
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20
5
E
15
D
15
C
25
B
40
A
Huffman Coding
35
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20
5
E
15
D
15
C
25
B
40
A
35
Huffman Coding
60
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20
5
E
15
D
15
C
25
B
40
A
35
Huffman Coding
60
1000
0
0
0
1
1
1
1
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Symbol Code
A 0
B 10
C 110
D 1111
E 1110
Consider the coding for ABCDE.
The code will be: 01011011111110
Can you decode it?
0 . 10 . 110. 1111 . 1110
Huffman Coding
Analysis: With the given frequencies, the expected number of bits
per character is: = 1X0.40 + 2X0.25+ 3X0.15 + 4X0.15 + 4 X 0.05
= 2.25
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